Question 820654: Hello, this is a question from my Algebra 2 textbook:
Find all values of c for which the equation has (a) two real solutions, (b) one real solution, and (c) two imaginary solutions.
x^2 - 2x + c = 0
Thanks!
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! In the case of x^2 - 2x + c = 0, it is of the form ax^2 + bx + c = 0 where a = 1, b = -2 and c is unknown.
The equation ax^2 + bx + c = 0 has at most 2 solutions. The type and number of unique solutions are determined by the equation below
D = b^2 - 4ac
which is the discriminant formula. The rules are this
* If D > 0, then you'll have 2 real solutions that are distinct.
* If D = 0, then you'll have exactly 1 real solution (that's also rational).
* If D < 0, then you'll have 2 imaginary solutions (aka, complex solutions)
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So from the rules above, we will have 2 real solutions when D > 0.
D > 0
b^2 - 4ac > 0 ... replace D with b^2 - 4ac (since D = b^2 - 4ac)
(-2)^2 - 4(1)c > 0 ... plug in a = 1, b = -2, and leave c alone
4 - 4c > 0
4 - 4c + 4c > 0 + 4c
4 > 4c
4c < 4
4c/4 < 4/4
c < 1
So if c < 1, then D > 0 which will lead to 2 distinct real solutions.
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Now let's move onto the second case: When D = 0
D = 0
b^2 - 4ac = 0
(-2)^2 - 4(1)c = 0
4 - 4c = 0
4 = 4c
4c = 4
c = 1
If c = 1, then you'll get exactly one real solution.
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You can probably guess what the last answer must be, but let's work through it
D < 0
b^2 - 4ac < 0
(-2)^2 - 4(1)c < 0
4 - 4c < 0
4 < 4c
4c > 4
c > 1
So if c > 1, then you'll get 2 imaginary solutions.
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